Given: int x / ( sinh(x) + cosh(x) ) dx∫xsinh(x)+cosh(x)dx
Multiply the integrand by 1 in the form of (sinh(x)-cosh(x))/(sinh(x)-cosh(x))sinh(x)−cosh(x)sinh(x)−cosh(x):
int x / ( sinh(x) + cosh(x) ) dx = int x / ( sinh(x) + cosh(x) )(sinh(x)-cosh(x))/(sinh(x)-cosh(x)) dx∫xsinh(x)+cosh(x)dx=∫xsinh(x)+cosh(x)sinh(x)−cosh(x)sinh(x)−cosh(x)dx
The denominator is multiplied using the difference of two squares pattern and the numerator is multiplied using the distributive property:
int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / ( sinh^2(x) - cosh^2(x)) dx∫xsinh(x)+cosh(x)dx=∫xsinh(x)−xcosh(x)sinh2(x)−cosh2(x)dx
The identity cosh^2(x)- sinh^2(x)=1cosh2(x)−sinh2(x)=1 tells us that the denominator is -1:
int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / -1 dx∫xsinh(x)+cosh(x)dx=∫xsinh(x)−xcosh(x)−1dx
Eliminate the denominator by changing signs in the numerator:
int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x)-xsinh(x) dx∫xsinh(x)+cosh(x)dx=∫xcosh(x)−xsinh(x)dx
Separate into two integrals:
int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x) dx- int xsinh(x) dx∫xsinh(x)+cosh(x)dx=∫xcosh(x)dx−∫xsinh(x)dx
Both integrals are trivial integrations by parts:
int x / ( sinh(x) + cosh(x) ) dx = (xsinh(x) - cosh(x)) - (xcosh(x)-sinh(x))+C∫xsinh(x)+cosh(x)dx=(xsinh(x)−cosh(x))−(xcosh(x)−sinh(x))+C
Regroup:
int x / ( sinh(x) + cosh(x) ) dx = xsinh(x)- xcosh(x)+ sinh(x)- cosh(x) +C∫xsinh(x)+cosh(x)dx=xsinh(x)−xcosh(x)+sinh(x)−cosh(x)+C